$\alpha^2 > 4\omega^2$, i.e., $\alpha > 2\omega$ (assuming $\omega$ is always positive). In this case $r_1, r_2 \in \mathbb{R}$ and the general solution to (2) is given by $y = A e^{r_1 t} + B e^{r_2 t}$. Indeed, both roots are negative so all solutions decay monotonically to zero as $t \to \infty$ (unless $\omega = 0$ in which case we have constant solutions).

Note that the greater of the two roots in case 1 is always greater than the root $-\alpha/2 = -\omega$ in case 2. We call case 1 overdamped because $\alpha$ is large and the greater root (i.e., less negative) leads to a slower exponential decay. In contrast, case 2 is called critically damped because solutions decay with time constant $\alpha/2 = \omega$, which is maximal. Case 3 is called underdamped because $\alpha$ is small and the damping force is insufficient to prevent oscillation. In this case the decay will also be slow since the time constant is $\alpha/2 < \omega$.

The solution to equation (3) is given by the sum of the general solution to the homogeneous equation (1) and a particular solution to (3) which we will solve by the method of undetermined coefficients. As hinted, let $y_p = A \cos (\beta t) + B \sin(\beta t)$. Then

By matching coefficients we find\begin{align}(-\beta^2 + \omega^2) A + \alpha\beta B &= 1 \\(-\alpha\beta) A + (-\beta^2 + \omega^2) B &= 0\end{align}This is a linear system in two equations and two unknowns, which is therefore easy to solve. The solution is\begin{align}A &= \frac{\omega^2 - \beta^2}{(\omega^2-\beta^2)^2 + (\alpha \beta)^2} \\B &= \frac{\alpha\beta}{(\omega^2 -\beta^2)^2 + (\alpha \beta)^2}\end{align}so the particular solution to (3) is $y_p = A \cos(\beta t) + B\sin(\beta t)$ with $A, B$ given above and the general solution to (3) is given by adding $y_p$ and the general solution to (1) as found above.